MATH 172: MOTIVATION FOR FOURIER SERIES: SEPARATION OF VARIABLES

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1 MATH 172: MOTIVATION FOR FOURIER SERIES: SEPARATION OF VARIABLES Separation of variabes is a method to sove certain PDEs which have a warped product structure. First, on R n, a inear PDE of order m is of the form a α (x α u = f(x, α m where a α, f are given functions on R n, and where we write and Typica exampes incude α = α αn n, α = α α n. (i Lapace s equation on domains Ω in R n : u = f, u = 2 1u nu, where u is the eectrostatic potentia generated by a charge distribution f, or the steady state temperature generated by a heat source f, (ii the wave equation on R t Ω, Ω a domain in R n, 2 t u c 2 x u = f, where c is the speed of waves, u is the dispacement of a membrane from equiibrium, or a component of the eectromagnetic fied, (iii and the heat equation on (0, t Ω t u k x u = f where k is (essentiay the heat conductivity, and u is the temperature generated by a heat source distribution f. In a these cases one needs to impose some boundary conditions if one is working in proper subdomains, and in the atter two some initia conditions. For instance, for the heat equation one may impose the Dirichet boundary condition (DBC u Ω = 0, resp. u R Ω = 0, resp. u (0, Ω = 0, representing an eectricay grounded or thermay insuating boundary for Lapace s equation, a fixed membrane edge for the wave equation, and a body whose surface is kept at 0 temperature for the heat equation. In addition, one woud need to specify the initia temperature u {0} Ω = φ for the heat equation, φ a given function on Ω, and the initia position and veocity for the wave equation u {0} Ω = φ, ( t u {0} Ω = ψ, 1

2 2 with φ, ψ given functions on Ω. Another common boundary condition is the Neumann boundary condition, u n Ω = 0, resp. u n R Ω = 0, resp. u n (0, Ω = 0, representing a thermay insuated boundary for Lapace s equation, a free membrane edge for the wave equation, and an body whose surface is insuated for the heat equation. In a these conditions, the 0 on the right hand side can be repaced by a specific function, e.g. for Lapace s equation, u Ω = h means that the boundary is kept at a fixed potentia h. The genera idea of separation of variabes is the foowing: suppose we have a inear PDE Lu = 0 on a space M x N y. We ook for soutions u(x, y = X(xY (y. In genera, there are no non-trivia soutions (the identicay 0 function being trivia, but in specia cases we might be abe to find some. We cannot expect even then that a soutions of the PDE are of this form. However, if we have a famiy u n (x, y = X n (xy n (y, n I, of separated soutions, where I is some index set (e.g. the positive integers, then, this being a inear PDE, u(x, y = n I c n u n (x, y = n I c n X n (xy n (y soves the PDE as we for any constants c n C, n I, provided the sum converges in some reasonabe sense, and we may be abe to choose the constants so that this in fact gives an arbitrary soution of the PDE. We emphasize that our endeavor, in genera, is very unreasonabe. Thus, we may make assumptions as we find it fit we need to justify our resuts after we derive them. As an exampe, consider the wave equation u tt c 2 x u = 0 on M x R t, where M is the space for instance, M is R n, or a cube [a, b] n or a ba B n = {x R n : x 1}. A separated soution is one of the form u(x, t = X(xT (t. Substituting into the PDE yieds X(xT (t c 2 T (t( x X(x = 0. Rearranging, and assuming T and X do not vanish, T (t c 2 T (t = xx(x. X(x Now, the eft hand side is a function independent of x, the right hand side is a function independent of t, so they are both equa to a constant, λ, namey pick your favorite vaue of x 0 and t 0, and then for any x and t, RHS(x = LHS(t 0 = RHS(x 0 = LHS(t, so the constant in question is LHS(t 0. Thus, we get two ODEs: T (t = λc 2 T (t, ( x X(x = λx(x.

3 3 Now typicay one has additiona conditions. For instance, one has boundary conditions at M: u M R = 0 (DBC or u n M R = 0 (NBC. Then X(xT (t has to satisfy these conditions for a x M and a t R. Taking some t for which T (t does not vanish, we deduce that the anaogous boundary condition is satisfied, namey X M = 0 (DBC or X n M = 0 (NBC. We aso have initia conditions, such as u(x, 0 = φ(x, u t (x, 0 = ψ(x, but as these are not homogeneous, we do not impose these at this point, and hope that we wi have sufficient fexibiity from the c n to match these. We start by soving the ODE for T, which is easy: T (t = A cos( λct + B sin( λct, λ 0, and T (t = A + Bt if λ = 0. (We coud have used compex exponentias instead. If λ is not positive, the trigonometric functions shoud be thought of as given by the corresponding compex exponentias. Now, in genera the spatia equation, X = λx, X M = 0 (DBC or (1 X n M = 0 (NBC is impossibe to sove expicity. However, we point out that it is an eigenvaue equation for : the statement is that X is an eigenfunction of with eigenvaue λ, in the strong sense that it aso satisfies the boundary condition. If we et X n (x, n N, be the eigenfunctions of with this boundary condition, with corresponding eigenvaue λ n, then the concusion is that u(x, t = n N A n cos( λ n ctx n (x + B n sin( λ n ctx n (x is the genera separated soution. Note that matching the initia conditions requires φ(x = n N A n X n (x, ψ(x = n N λn cb n X n (x, i.e. writing the given functions φ and ψ as infinite inear combination of the eigenfunctions X n of course, in addition to finding some A n and B n which shoud work, we need to actuay prove that these series indeed converges to the desired imit. Due to the decomposition of u into eigenfunctions of the spatia operator, X, these methods for soving the PDE are aso caed spectra methods. In a simpe situation, such as when M = [0, ], we can find the eigenfunctions of expicity. Namey, for Dirichet boundary condition, the equation is d2 X dx 2 = λx, X(0 = 0 = X(,

4 4 the soution of the ODE (without the boundary conditions is (2 X(x = C cos( λx + D sin( λx, λ 0, X(x = C + Dx, λ = 0. Evauating at x = 0 and enforcing X(0 = 0 yieds C = 0 in either case. Evauating at yieds D = 0 if λ = 0, so λ = 0 is of no interest (we are ony interested in non-trivia soutions. If λ 0, then evauation at x = yieds that either D = 0, which again woud give a trivia soution, or sin( λ = 0, which thus we assume is the case. But the zeros of sine are at nπ, n Z, so λ = nπ, hence ( nπ λ =. Note that n and n give essentiay the same X n (up to an overa minus sign, which is irreevant, as we are aowing an arbitrary coefficent D, whie n = 0 yieds the trivia soution, so we concude that X n (x = sin, λ n =, n > 0, n Z. Returning to the fu probem, we deduce that u(x, t = A n cos( λ n ct sin + B n sin( λ n ct sin. It remains to determine the coefficients A n and B n, which is a subject of the next ecture. Next, we consider the Neumann boundary conditions. d2 X dx 2 = λx, X (0 = 0 = X (, Then (2 sti hods. Substituting in X (0 = 0 yieds D = 0 whether λ = 0 or not. If λ = 0, the constant C satisfies the boundary condition at, so 1 is an eigenfunction with eigenvaue 0. If λ 0, we obtain the requirement (as we want a non-trivia soution λ sin( λ = 0, hence λ = nπ, n Z, as above. Since λ 0, we in fact have n 0 in this case. Again, n and n give essentiay the same eigenfunctions (in fact, exacty!, so our eigenfunctions are X n (x = cos, λ n =, n > 0, n Z X 0 (x = 1, λ 0 = 0. Note that n = 0 can be considered simpy a specia case of the genera formua in this particuar situation, and thus we may write our answer as X n (x = cos, λ n =, n 0, n Z, hence the genera separated soution as u(x, t = A 0 + B 0 t + A n cos( λ n ct cos + B n sin( λ n ct cos.

5 5 Next, consider the heat equation on a space M, i.e. u t k x u = 0 with either Dirichet or Neumann boundary conditions. u(x, t = X(xT (t we obtain so X(xT (t kt (t x X = 0, kt = xx X. Again, both sides must be equa to a constant, λ. The ODE for T, is then easy to sove, T T = λkt, T (t = Ae λkt. The PDE for X sti has boundary conditions, and is (3 X = λx, X M = 0 (DBC or X n M = 0 (NBC, For separated soutions the same as (1. The soution is thus aso the same. For instance, for Dirichet boundary conditions on [0, ], X n (x = sin, λ n =, n > 0, n Z, hence the genera separated soution is u(x, t = A n e kλnt sin. Simiary, for Neumann boundary conditions we obtain u(x, t = A n e kλnt cos. n=0 Simiar techniques aso work for some other probems with ess product-type behavior. Thus, consider Lapace s equation on the disk of radius R, namey D = B 2 R = {x R 2 : x < R}, u = 0, x D, u D = h, with h a given function on D. We again ignore the inhomogeneous conditions, so we are eft with the PDE. Now, we need to think of the disk as a product space. This is not the case in Cartesian coordinates. However, in poar coordinates, we can identify D with [0, R r S 1 θ, where S 1 is the unit circe. The Lapacian in poar coordinates is = 2 r + r 1 r + r 2 2 θ.

6 6 Note that poar coordinates are singuar at r = 0 (the whoe circe at r = 0 is squashed into a singe point, and has coefficients which are not smooth at r = 0. We again consider separated soutions, Substituting into the PDE yieds Dividing by ΘR and mutipying by r 2 : u(r, θ = R(rΘ(θ. R Θ + r 1 R Θ + r 2 Θ R = 0. r2 R + rr = Θ R Θ, and as before both sides must be equa to a constant, λ. Thus, the Θ ODE is Θ = λθ, where Θ is a function on the circe S 1. A function on S 1 can be thought of as a function on [0, 2π] such that the vaues at 0 and 2π agree (since they represent the same point on S 1. However, a better way of thinking of it as a 2π-periodic function on R: the points θ + 2nπ, n Z, correspond to the same point in the circe. Expicity, if the circe is considered the unit circe in R 2, as usua, the map R S 1 is R θ (cos θ, sin θ S 1, making the 2π-periodicity cear. Thus, we want soutions of the Θ ODE, considered as an ODE on the rea ine, which are 2π-periodic. Now, without the periodicity condition, the soution of the ODE is, as before (when this was the ODE for T, and Θ(θ = A cos( λθ + B sin( λθ, λ 0, Θ(θ = A + Bθ if λ = 0. Adding the 2π periodicity condition we need that λ = n Z if λ 0, and B = 0 if λ = 0. If λ 0, then this gives λ = n 2. As positive and negative vaues of n give rise to the same eigenfunctions, we can restrict to n > 0 (note n 0 if λ 0. Thus, in summary, the soutions are Θ n (θ = A n cos(nθ + B n sin(nθ, n > 0, n Z, Θ 0 (θ = A 0. It remains to dea with the R ODE which is r 2 R + rr λr = 0. Note that the coefficient of the highest derivative, R 2, vanishes at r = 0, so this ODE is degenerate, or singuar, there. However, note that each derivative comes with a factor of r, thus R (one derivative faing on R has a factor of r in the coefficient, R (two derivatives faing on R has a factor of r 2. A sighty better way to rewrite it is r(rr λr = 0, then expicity you can see that each time you differentiate, you mutipy by r. Such an ODE is caed a reguar singuar ODE (in genera, one coud mutipy by a smooth function times r each times one differentiates, and can (usuay be soved in a power series (in genera with non-integer powers around r = 0, with possibe ogarithmic terms. Our ODE is a particuary simpe reguar singuar ODE. This

7 can be seen by changing variabes t = og r, writing T (r = og r, so dt dr = r 1, for then for a function f on R, r d(f T ( df dt = r dr dt T dr = df dt T, or informay, r df dr = r df dt dt dr = df dt, so the ODE for F (t = R(e t becomes so d 2 F λf = 0, dt2 F (t = Ae λt + Be λt, λ 0, F (t = A + Bt, λ = 0. Using t = og r, we get for the origina function R, R(r = Ar λ + Br λ, λ 0, R(r = A + B og r, λ = 0. Recaing that λ is a non-negative integer from the Θ ODE, we concude that R(r = Ar n + Br n, n > 0, n Z, R(r = A + B og r, n = 0. Now, there is nothing specia about the origin: it is just a point in the interior of the disk, and we know by eiptic reguarity that the soution of Lapace s equation shoud be C in the disk, so in particuar it shoud be bounded, hence we throw out the exponents that woud yied unbounded terms to get R n (r = r n, n > 0, n Z, R 0 (r = 1, n = 0, and the n = 0 case coud be simpy incuded in the n > 0 one by etting n 0 there. Note that the bady behaved soutions r n and og r arose because we used bady behaved coordinates on D: reca that these themseves were singuar at the origin. Combining this with our resuts for Θ, we obtain the genera separated soution u(r, θ = A 0 + r n (A n cos(nθ + B n sin(nθ. Again, the question is whether we can determine the coefficients to match the boundary conditions. Expicity, the boundary condition is h(θ = A 0 + R n (A n cos(nθ + B n sin(nθ, so we need to write the given function h as an infinite inear combinations of the sines and cosines, and discuss convergence of the resut. This is the topic of the next ectures. 7